Question If the closed loop transfer function of a system is G(s) K(s+1) s(s+a) (2+s+1) and for K=6 and a=4, which one is true related with Routh-Hurwitz stability analysis of the system. A. There are no sign changes in the first collumn of Routh-Hurwitz Array. B. System is stable. C. There are two sign changes in the first collumn of Routh-Hurwitz Array. D. System in unstable. E. A and B true. F. Cand D true. G. All answers are wrong.

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Transcribed Image Text: If the closed loop transfer function of a system is G(s) K(s+1) s(s+a) (2+s+1) and for K=6 and a=4, which one is true related with Routh-Hurwitz stability analysis of the system. A. There are no sign changes in the first collumn of Routh-Hurwitz Array. B. System is stable. C. There are two sign changes in the first collumn of Routh-Hurwitz Array. D. System in unstable. E. A and B true. F. Cand D true. G. All answers are wrong.

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G(s)=\frac{K(s+1)}{s(s+a)\left(s^{2}+s+1\right)} \quad k=6, a=4Charactaristic equation (E)=1+G(s) H(s)=0\begin{array}{l}1+\frac{k(s+1)}{s(s+a)\left(s^{2}+s+1\right)}=0 \\C \cdot E=s(s+4)\left(s^{2}+s+1\right)+6(s+1)=0 \\s^{4}+5 s^{3}+5 s^{2}+4 s+6 s+6=0 \\s^{4}+5 s^{3}+5 s^{2}+10 s+6=0 \rightarrow \text { No-sign change } \\\end{array}By R=H creteria coption (-wrong)Aurilary equation \Rightarrow 3 s^{2}+6=0=A E\begin{array}{l}\frac{d(A E)}{d s}=0 \Rightarrow 3 \frac{d}{d s} s^{2}+\frac{d}{d s}(\sigma)=3(2 s)+0 \\\frac{d A E)}{d s}=6 s+0 \\\end{array}Anilang Erration 2\begin{array}{l}\text { Imginayy pule }=j \omega=2 \\\operatorname{RHP}(\text { Rigut hand poles) }=0 \rightarrow \text { system is stable } \\\operatorname{LHP}(\text { Lett hand poles) }=2\end{array}No-singn Chang in R+ tabulation \rightarrow option (a) cosrect\rightarrow system is marginallystable because two potes on inaginary axis & No-Righ hand poles, system is stable option (b)So: option E \rightarrow both A \varepsilon B is true-correct correct ...